Quick Calculation 3.11

A First Course in String Theory

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Since $\displaystyle{G^{(D)} \rho_m}$ has the same unit in all dimensions,

\displaystyle{ \begin{aligned} \left[ G^{(D)} {\rho_m}_D \right] &= \left[ G^{(D=4)} {\rho_m}_{D=4} \right] \\ \left[ G^{(D)} \right] \frac{M}{L^{D-1}} &= \left[ G \right] \frac{M}{L^3} \\ \left[ G^{(D)} \right] &= \left[ G \right] L^{D-4} \\ \end{aligned} }

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Eq. (3.104):

\displaystyle{ \begin{aligned} [G] &= \frac{[c]^3 L^2}{[\hbar]} \\ \end{aligned}}

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\displaystyle{\begin{aligned} \left[ G^{(D)} \right] &= \frac{[c]^3 L^{D-2}}{[\hbar]} \\ G^{(D)} &= \frac{c^3 \left(l_P^{(D)}\right)^{D-2}}{\hbar} \\ \left(l_P^{(D)}\right)^{D-2} &= G^{(D)} \frac{\hbar}{c^3} \\ \end{aligned}}

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\displaystyle{\begin{aligned} \left(l_P^{(4)}\right)^{4-2} &= G^{(4)} \frac{\hbar}{c^3} \\ \left(l_P \right)^{2} &= G \frac{\hbar}{c^3} \\ \\ \left(l_P^{(D)}\right)^{D-2} &= \left(l_P \right)^{2} \frac{G^{(D)}}{ G } \\ \\ \end{aligned}}

— Me@2022-07-17 04:23:42 PM

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